College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.6 - Solving Other Types of Equations - 1.6 Exercises - Page 138: 13


$x=0$ or $x=-2$

Work Step by Step

We solve by factoring: $x^{5}+8x^{2}=0$ $x^{2}(x^{3}+8)=0$ $x^{2}(x+2)(x^{2}-2x+4)=0$ $x^{2}=0$ or $x+2=0$ or $x^{2}-2x+4=0$ $x=0$ or $x=-2$ (The last equation $x^{2}-2x+4=0$ has no real solutions because the discriminant is negative.)
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